A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. 'm' grams of steam at 100°C

Q: What exactly is 'water equivalent'?

The water equivalent of a body is the mass of water that would absorb or release the same amount of heat as the body for the same change in temperature. It allows us to treat the calorimeter simply as an extra $20\text{ g}$ of water, making calculations much easier.

Q: Why do we add the water equivalent to the mass of water?

Since both the water and the calorimeter start at the same initial temperature ($25^\circ\text{C}$) and reach the same final temperature ($31^\circ\text{C}$), they undergo the exact same temperature change. Adding their effective masses allows us to calculate the total heat gained in one single step.

Q: Why does steam release so much heat compared to hot water?

Steam possesses 'latent heat of vaporization'. This is the massive amount of energy that was required to break the intermolecular bonds of liquid water during boiling. When steam condenses back into water, all this stored energy is released at once, which is why $540\text{ cal/g}$ is so much larger than the specific heat of $1\text{ cal/g}^\circ\text{C}$.

Q: Do we need to convert grams to kilograms or Celsius to Kelvin?

Not necessarily. As long as you use a consistent system of units, you are fine. Here, we are using the CGS system (calories, grams, degrees Celsius). Since all given values ($L_v$, $s_w$) are in CGS, keeping the mass in grams and temperature in Celsius works perfectly and avoids messy conversions.

Q: What if the final temperature was $100^\circ\text{C}$?

If the final equilibrium temperature was $100^\circ\text{C}$, the steam would only condense into water at $100^\circ\text{C}$ and would not cool down further. In that case, the heat lost would simply be $m \cdot L_v$, and the $Q_2$ term (cooling) would be zero.

Question

A calorimeter of water equivalent $20	ext{ g}$ contains $180	ext{ g}$ of water at $25^\circ	ext{C}$. '$m$' grams of steam at $100^\circ	ext{C}$ is mixed in it till the temperature of the mixture is $31^\circ	ext{C}$. The value of $m$ is close to
(Take, latent heat of water $= 540	ext{ cal g}^{-1}$, specific heat of water $=1	ext{ cal g}^{-1}\ ^\circ	ext{C}^{-1}$)

Options:

(A) $2	ext{ g}$

(B) $3.2	ext{ g}$

(C) $2.6	ext{ g}$

(D) $4	ext{ g}$

Sigma Percentile · Accepted Answer

### The Dance of Heat: Steam Meets Water

Imagine a cold winter morning. You have a cup of water at $25^\circ	ext{C}$, and you decide to warm it up by blasting $100^\circ	ext{C}$ steam into it. This is exactly what our problem is about. We are mixing a hot substance (steam) with a cold substance (water and its container) until they reach a cozy equilibrium at $31^\circ	ext{C}$.

### The Principle of Calorimetry

The core principle governing this entire process is the conservation of energy, beautifully packaged as the **Principle of Calorimetry**. It states a simple truth: assuming no heat escapes into the surrounding air, the heat lost by the hot bodies must perfectly equal the heat gained by the cold bodies.

$$Q_{	ext{lost}} = Q_{	ext{gained}}$$

### Analyzing the Cold Side

Let's look at the cold side of our system first. We have $180	ext{ g}$ of water sitting in a calorimeter. The problem gives us a neat little trick: the calorimeter has a **water equivalent** of $20	ext{ g}$. What does this mean? It means the material of the calorimeter absorbs heat exactly like $20	ext{ g}$ of water would.

Instead of calculating the heat gained by the water and the calorimeter separately, we can combine them into a single, effective mass of water:

$$M_{	ext{total}} = 180	ext{ g} + 20	ext{ g} = 200	ext{ g}$$

This $200	ext{ g}$ of effective water warms up from $25^\circ	ext{C}$ to $31^\circ	ext{C}$. The heat gained is calculated using the specific heat formula $Q = ms\Delta T$:

$$Q_{	ext{gained}} = 200 	imes 1 	imes (31 - 25)$$
$$Q_{	ext{gained}} = 200 	imes 6 = 1200	ext{ cal}$$

### The Journey of the Steam

Now, let's trace the journey of the steam. The steam doesn't just cool down; it undergoes a dramatic phase change. This happens in two distinct steps.

**Step 1: Condensation**
The steam at $100^\circ	ext{C}$ must first condense into liquid water at $100^\circ	ext{C}$. This phase change releases a massive amount of energy known as the latent heat of vaporization ($L_v = 540	ext{ cal/g}$).

$$Q_1 = m 	imes L_v = m 	imes 540$$

**Step 2: Cooling**
Now we have $m$ grams of hot water at $100^\circ	ext{C}$. This water must cool down to the final equilibrium temperature of $31^\circ	ext{C}$. This releases sensible heat.

$$Q_2 = m 	imes s_w 	imes (100 - 31) = m 	imes 1 	imes 69 = 69m$$

The total heat lost by the steam is the sum of these two processes:

$$Q_{	ext{lost}} = 540m + 69m = 609m	ext{ cal}$$

### The Final Calculation

We bring it all together by equating the heat lost to the heat gained:

$$609m = 1200$$

Solving for $m$:

$$m = \frac{1200}{609} \approx 1.97	ext{ g}$$

This value is incredibly close to **$2	ext{ g}$**.

Take a moment to appreciate this result. It took merely $2$ grams of steam to heat up $200$ grams of water by $6$ degrees! This perfectly illustrates the immense power of latent heat. The energy required to break the bonds of liquid water to form steam is vast, and all of that energy is returned when the steam condenses back into water.

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Visualized Solution

$System Setup$

$Principle of Calorimetry$

$Heat Gained by Cold System$

$Calculating Q_{gained}$

$Heat Lost by Steam$

$Total Heat Lost$

$Equating and Solving$

$The Power of Latent Heat$

The Sigma Insight: Calorimetry

The Dance of Heat

The Principle of Calorimetry

Analyzing the Cold Side

The Journey of the Steam

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Principle of Calorimetry

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Gained by Cold System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Qgained​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Lost by Steam

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Heat Lost

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating and Solving

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Power of Latent Heat